University Physics IV · topic UPIV-08.06
Eigenfunctions, nodes, and the classically forbidden region
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Eigenfunctions, nodes and the forbidden region: predict a finite change
The oscillator eigenfunctions are Hermite-polynomial-times-Gaussian states with n nodes and parity (−1)ⁿ, orthogonal across the ladder, frozen in time, and extending measurably past the classical turning points ±√(2n+1)x₀. Use the supplied topic model Y_out = (p^2-(2*n+1))/2, where Y_out is V − E AT PROBE (ħω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. p is Probe position ξ, measured in x₀. Keep n (Quantum number n) = 0 fixed. For this exercise use only -2.835 x₀ ≤ p ≤ -0.189 x₀ on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at p = -2.25288 x₀ and -1.88244 x₀. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δp across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
V − E AT PROBE against p
Predictions from the supplied Eigenfunctions, nodes and the forbidden region model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): V − E AT PROBE / ħω
Horizontal axis (x): p: Probe position ξ / x₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Probe position ξ / x₀ | V − E AT PROBE / ħω |
|---|---|
| -2.835 | 3.5186124999999997 |
| -2.1734999999999998 | 1.8620511249999994 |
| -1.512 | 0.6430720000000001 |
| -0.8505 | -0.13832487499999996 |
| -0.18900000000000006 | -0.4821395 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of p separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At p = -2.25288 x₀, substitute into Y_out = (p^2-(2*n+1))/2: Y₀ = 2.03773 ħω.
- At p = -1.88244 x₀, the same fixed-parameter model gives Y₁ = 1.27179 ħω.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.765944 ħω; the input change is Δp = 0.37044 x₀.
- Average sensitivity is ΔY_out/Δp = -2.06766 ħω per x₀. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 2.03773 ħω, Y₁ = 1.27179 ħω; ΔY_out = -0.765944 ħω; average sensitivity = -2.06766 ħω per x₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Eigenfunctions, nodes and the forbidden region: infer an input with uncertainty
The oscillator eigenfunctions are Hermite-polynomial-times-Gaussian states with n nodes and parity (−1)ⁿ, orthogonal across the ladder, frozen in time, and extending measurably past the classical turning points ±√(2n+1)x₀. Use the supplied topic model Y_out = (p^2-(2*n+1))/2, where Y_out is V − E AT PROBE (ħω). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. p is Probe position ξ, measured in x₀. Keep n (Quantum number n) = 0 fixed. For this exercise use only -2.835 x₀ ≤ p ≤ -0.189 x₀ on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.24534724 ħω, with a bounded reading uncertainty ±0.10912 ħω. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0002646 x₀, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
V − E AT PROBE against p
Predictions from the supplied Eigenfunctions, nodes and the forbidden region model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): V − E AT PROBE / ħω
Horizontal axis (x): p: Probe position ξ / x₀
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Probe position ξ / x₀ | V − E AT PROBE / ħω |
|---|---|
| -2.835 | 3.5186124999999997 |
| -2.1734999999999998 | 1.8620511249999994 |
| -1.512 | 0.6430720000000001 |
| -0.8505 | -0.13832487499999996 |
| -0.18900000000000006 | -0.4821395 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(p) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(-2.835) = 3.51861 ħω and Y_out(-0.189) = -0.482139 ħω. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives p_line = -2.835 + (0.245347 − 3.51861)(-0.189 − -2.835)/(-0.482139 − 3.51861) = -0.670142 x₀. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives p = -1.22094 x₀; substituting back gives Y_out = 0.24534724 ħω.
- The interpolation error in the input is |p_line − p| = 0.550798 x₀. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.13622724 ħω and 0.35446724 ħω. Sort the roots even if the response decreases: -1.30726 x₀ ≤ p ≤ -1.12803 x₀.
- The lower and upper input deviations are 0.0863222 x₀ and 0.0929088 x₀. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer p ≈ -1.22094 x₀; compatible interval [-1.30726, -1.12803] x₀. Endpoint interpolation gives -0.670142 x₀.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.